MHT CET · Chemistry · Thermodynamics (C)
Calculate the work done in the oxidation of one mole \(\mathrm{HCl}_{(\mathrm{g})}\) at \(27^{\circ} \mathrm{C}\), according to reaction.
\(4 \mathrm{HCl}_{(\mathrm{g})}+\mathrm{O}_{2_{(\mathrm{g})}} \rightarrow 2 \mathrm{Cl}_{2_{(\mathrm{g})}}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \quad\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A 2494.2 J
- B 623.6 J
- C \(1247 \cdot 1 \mathrm{~J}\)
- D 1870.7 J
Answer & Solution
Correct Answer
(B) 623.6 J
Step-by-step Solution
Detailed explanation
\(\Delta n_g = (2+2) - (4+1) = -1\) For 1 mole HCl: \(\Delta n_g = \frac{-1}{4}\)
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